We show the validity of OPs claim by using some operator calculations and deriving binomial identities. We start with some notational conventions. Here we use rising and falling factorial powers
\begin{align*}
x^{\overline{n}}&=x(x+1)\cdots(x+n-1)\\
x^{\underline{n}}&=x(x-1)\cdots(x-n+1)\\
\end{align*}
and the differential operator $D_x=\frac{d}{dx}$. Using this notation the operator $\hat{\mathcal{O}_n}$ can be written $$\hat{\mathcal{O}_n}=\frac{1}{n!}(xD_x+n)^{\underline{n}}$$
We now formulate OPs claim in two steps. At first we transform the operator to a more convenient representation which will be used in the proof below.
The following is valid
\begin{align*}
\hat{\mathcal{O}_n}=\frac{1}{n!}(xD_x+n)^{\underline{n}}
=\sum_{k=0}^n\binom{n}{k}\frac{1}{k!}x^kD_x^k\qquad &\qquad n\geq 1\tag{1}\\
\end{align*}
and then we apply the operator to $\frac{\log x}{x^m}$
\begin{align*}
\hat{\mathcal{O}_n}\left(\frac{\log x}{x^m}\right)&=\frac{1}{n!}(xD_x+n)^{\underline{n}}\left(\frac{\log x}{x^m}\right)\\
&=\frac{(-1)^{m+1}}{mx^m}\binom{n}{m}^{-1}\qquad &\qquad 1\leq m\leq n\tag{2}
\end{align*}
Approach: The proof is divided into four steps. At first we derive the operator representation (1). Then we apply (1) to $\frac{\log x}{x^m}$ and derive an expression containing a sum with $x^{-m}\log x$ times a binomial expression and another sum with $x^{-m}$ times a binomial expression. In step three we show the first binomial sum cancels and in the last step we simplify the other sum to obtain (2).
Operator methods:
We obtain
\begin{align*}
\hat{\mathcal{O}_n}&=\frac{1}{n!}(xD_x+n)^{\underline{n}}\tag{3}\\
&=\frac{1}{n!}\sum_{k=0}^n\binom{n}{k}\left(xD_x\right)^{\underline{k}}n^{\underline{n-k}}\\
&=\sum_{k=0}^n\binom{n}{k}\frac{1}{k!}\left(xD_x\right)^{\underline{k}}\tag{4}\\
&=\sum_{k=0}^n\binom{n}{k}\frac{1}{k!}\sum_{j=0}^k(-1)^{k-j}\left[k\atop j\right]\left(xD_x\right)^j\tag{5}\\
&=\sum_{k=0}^n\binom{n}{k}\frac{1}{k!}\sum_{j=0}^k(-1)^{k-j}\left[k\atop j\right]\sum_{l=0}^j{j\brace l}x^lD_x^l\tag{6}\\
&=\sum_{k=0}^n\binom{n}{k}\frac{1}{k!}x^kD_x^k\tag{7}
\end{align*}
and the claim (1) follows.
Comment:
In (3) we use the binomial theorem which is also valid for falling factorials. See e.g. exercise 5.37 in Concrete Mathematics by D.E. Knuth et al.
In (4) we use the representation of falling factorial powers by Stirling numbers of the first kind $\left[n\atop k\right]$
\begin{align*}
x^{\underline{n}}=\sum_{k=0}^n(-1)^{n-k}\left[n\atop k\right]x^j
\end{align*}
In (5) we use the representation of $\left(xD_x\right)^n$ by Stirling numbers of the second kind ${n\brace k}$
\begin{align*}
\left(xD_x\right)^n=\sum_{k=0}^n(-1)^{n-k}{n\brace k}x^kD_x^k
\end{align*}
In (6) we use that Stirling numbers of first and second kind are binomial inverse pairs.
\begin{align*}
\sum_{k=0}^n(-1)^{n-k}\left[n\atop k\right]{k\brace l}=[[n=l]]
\end{align*}
with $[[P]]$ the Iverson bracket notation.
Intermezzo:
According to the transformation (7) of the operator $\hat{\mathcal{O}_n}$ we next calculate up the $k$-th derivative of $\frac{\log x}{x^m}$. We do this by applying the general Leibniz rule
\begin{align*}
D_x^k\left(\frac{\log x}{x^m}\right)&=\sum_{j=0}^k\binom{k}{j}D_x^j\left(\log x\right)D_x^{k-j}\left(x^{-m}\right)\\
&=\log x D_x^k\left(x^{-m}\right)+\sum_{j=1}^k\binom{k}{j}D_x^j\left(\log x\right)D_x^{k-j}\left(x^{-m}\right)
\end{align*}
Since
\begin{align*}
D_x^k\left(x^{-m}\right)&=(-1)^km^{\overline{k}}x^{-m-k}\\
D_x^k(\log x)&=(-1)^{k-1}(k-1)!x^{-k}
\end{align*}
we obtain
\begin{align*}
D_x^k\left(\frac{\log x}{x^m}\right)&=(-1)^km^{\overline{k}}x^{-m-k}\log x\\
&\qquad -(-1)^kx^{-m-k}\sum_{j=1}^k\binom{k}{j}(j-1)!m^{\overline{k-j}}\qquad\qquad m,k\geq 1
\end{align*}
$$ $$
Applying the operator:
Combining the last result with (7) we obtain after some simplifications
\begin{align*}
\frac{1}{n!}&(xD_x+n)^{\underline{n}}\left(\frac{\log x}{x^m}\right)\\
&=\sum_{k=0}^n\binom{n}{k}\frac{1}{k!}x^kD_x^k\left(\frac{\log x}{x^m}\right)\\
&=x^{-m}\log x\sum_{k=0}^m\binom{n}{k}(-1)^k\frac{1}{k!}m^{\overline{k}}\\
&\qquad-x^{-m}\sum_{k=1}^n\binom{n}{k}(-1)^k\frac{1}{k!}\sum_{j=1}^k\binom{k}{j}(j-1)!m^{\overline{k-j}}\tag{8}
\end{align*}
The last steps consist of simplification of both sums in the last expression (8). In fact this simplification provides some nice binomial identities.
Cancelling first binomial sum:
The following is valid
\begin{align*}
\sum_{k=0}^n\binom{n}{k}(-1)^k\frac{1}{k!}m^{\overline{k}}=0\qquad\qquad m,n\geq 1\tag{9}
\end{align*}
We get
\begin{align*}
\sum_{k=0}^n\binom{n}{k}(-1)^k\frac{1}{k!}m^{\overline{k}}
&=\sum_{k=0}^n(-1)^k\binom{m+k-1}{k}\tag{10}\\
&=\sum_{k=0}^n\binom{n}{k}\binom{-m}{k}\tag{11}\\
&=\binom{n-m}{n}\\
&=[[n=m=0]]
\end{align*}
and (9) follows.
Comment:
Simplifying second binomial sum:
The following is valid
\begin{align*}
\sum_{k=1}^n\binom{n}{k}\sum_{j=1}^k\frac{1}{j}(-1)^j\binom{-m}{k-j}
&=\sum_{j=0}^{n-m}\frac{1}{j+m}(-1)^{j+m}\binom{n-m}{j}\tag{12}\\
&=\frac{(-1)^m}{m}\binom{n}{m}^{-1} \qquad\qquad1\leq m\leq n\tag{13}
\end{align*}
We show the identity (12) by application of the Chu-Vandermonde identity. The step from (12) to (13) is a nice identity by its own. We will show it using operator methods.
We obtain
\begin{align*}
\sum_{k=1}^n&\binom{n}{k}\sum_{j=1}^k\frac{1}{j}(-1)^j\binom{-m}{k-j}\tag{14}\\
&=\sum_{j=1}^n\sum_{k=j}^n\frac{1}{j}(-1)^j\binom{n}{k}\binom{-m}{k-j}\tag{15}\\
&=\sum_{j=1}^n\frac{1}{j}(-1)^{j}\sum_{k=0}^{n-j}\binom{n}{k+j}\binom{-m}{k}\tag{16}\\
&=\sum_{j=m}^n\frac{1}{j}(-1)^{j}\binom{n-m}{n-j}\tag{17}\\
&=\sum_{j=0}^{n-m}\frac{1}{m+j}(-1)^{m+j}\binom{n-m}{j}
\end{align*}
and (10) follows.
Comment:
In (14) we exchange the sums.
In (15) we shift the index $k$ of the inner sum to start from $k=0$.
In (16) we apply the Chu-Vandermonde identity.
In (17) we shift the index $j$ to start from $j=0$.
Note the shift operator $E$ with $E(f(m))=f(m+1)$ and the delta operator $\Delta:=E-1$ are related by
\begin{align*}
\Delta^n=(E-1)^n=\sum_{j=0}^n\binom{n}{j}(-1)^{n-j}E^j\tag{17}
\end{align*}
applying the delta operator to $f(m)=\frac{1}{m}$ results in
\begin{align*}
\Delta\left(\frac{1}{m}\right)&=\frac{1}{m+1}-\frac{1}{m}=-\frac{1}{m(m+1)}\\
\Delta^2\left(\frac{1}{m}\right)&=-\frac{1}{(m+1)(m+2)}-\frac{1}{m(m+1)}=\frac{2}{m(m+1)(m+2)}\\
&\cdots\\
\Delta^n\left(\frac{1}{m}\right)&=\frac{(-1)^nn!}{m(m+1)\cdots(m+n)}=\frac{(-1)^n}{m}\binom{m+n}{n}^{-1}\tag{18}
\end{align*}
which can be shown easily by using mathematical induction.
Applying (17) and (18) is essentially the proof of (11) since
\begin{align*}
\Delta^n\left(\frac{1}{m}\right)&=\sum_{j=0}^n\binom{n}{j}(-1)^{n-j}E^j\left(\frac{1}{m}\right)\\
&=\sum_{j=0}^n\binom{n}{j}(-1)^{n-j}\frac{1}{m+j}\\
&=\frac{(-1)^m}{m}\binom{m+n}{n}^{-1}
\end{align*}
Replacing $n$ with $n-m$, cancelling $(-1)^n$ and the claim (11) follows.
Finally putting all together
we obtain from (8), (9) and (13)
\begin{align*}
\frac{1}{n!}(xD_x+n)^{\underline{n}}\left(\frac{\log x}{x^m}\right)
&=\sum_{k=0}^n\binom{n}{k}\frac{1}{k!}x^kD_x^k\left(\frac{\log x}{x^m}\right)\\
&=x^{-m}\log x\sum_{k=0}^m\binom{n}{k}(-1)^k\frac{1}{k!}m^{\overline{k}}\\
&\qquad-x^{-m}\sum_{k=1}^n\binom{n}{k}(-1)^k\frac{1}{k!}\sum_{j=1}^k\binom{k}{j}(j-1)!m^{\overline{k-j}}\\
&=x^{-m}\log x\cdot 0-x^{-m}\cdot\frac{(-1)^m}{m}\binom{n}{m}^{-1}\\
&=\frac{(-1)^{m+1}}{mx^m}\binom{n}{m}^{-1}\qquad \qquad 1\leq m\leq n\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}
and the claim (2) follows.
